题目内容
定义运算m⊙n=m2-mn-n2,则cos
⊙sin
=( )
| π |
| 6 |
| π |
| 6 |
分析:依题意,cos
⊙sin
=cos2
-cos
•sin
-sin2
,利用二倍角公式即可求得答案.
| π |
| 6 |
| π |
| 6 |
| π |
| 6 |
| π |
| 6 |
| π |
| 6 |
| π |
| 6 |
解答:解:∵m⊙n=m2-mn-n2,
∴cos
⊙sin
=cos2
-cos
•sin
-sin2
=cos
-
sin
=
-
.
故选A.
∴cos
| π |
| 6 |
| π |
| 6 |
=cos2
| π |
| 6 |
| π |
| 6 |
| π |
| 6 |
| π |
| 6 |
=cos
| π |
| 3 |
| 1 |
| 2 |
| π |
| 3 |
=
| 1 |
| 2 |
| ||
| 4 |
故选A.
点评:本题考查三角函数的恒等变换及化简求值,着重考查二倍角的正弦与余弦公式,属于中档题.
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